VITEEE PCBE Mock Test - 5 - JEE MCQ

The key reason viruses were not considered living organisms in Whittaker's classification is their lack of cellular structure, rendering them acellular. Unlike living cells, viruses must infect host cells to replicate, as they cannot carry out metabolic processes on their own. This is why Option B is correct. The other options are incorrect: Viruses can reproduce but only inside a host cell (Option C), they are not necessarily larger than bacteria (Option A), and they contain either DNA or RNA, but not both (Option D).

Viruses lack cellular structure, making them non-living outside hosts, so they’re excluded from Whittaker’s classification.

Ascomycetes, commonly known as sac-fungi, are distinguished by their mode of sexual reproduction, where they produce ascospores endogenously within sac-like structures called asci. These asci are often organized into larger fruiting bodies known as ascocarps. This makes Option C the correct answer. Option A is incorrect as zoospores are characteristic of some other fungi groups like Phycomycetes. Option B is also incorrect for Ascomycetes, whose mycelium is typically septate and not coenocytic. Option D is incorrect as it describes Deuteromycetes, not Ascomycetes.

Ascomycetes produce ascospores in asci, a hallmark of their sexual reproduction.

Answer: A. Chitinous exoskeleton

Solution:
Arthropods, which include insects, arachnids, crustaceans, and myriapods, are characterized by their chitinous exoskeleton, jointed limbs, and segmented bodies. The presence of a notochord is unique to Chordata, radial symmetry is a feature of Cnidaria and Echinodermata (in their larval stage), and the water vascular system is specific to Echinodermata

For a dividing bacterium:
A: Cell wall (outer layer).
B: Cell membrane (beneath wall).
C: DNA (genetic material).
D: Heterocyst (not typical in most bacteria, possibly a typo for nucleoid).
E: Mucilaginous sheath (external layer in some bacteria).
Correct: A - Cell wall, B - Cell membrane, C - DNA, D - Heterocyst (if specific), E - Mucilaginous sheath.

Wolf and Tasmania Wolf are examples of convergent evolution of placental mammals and Australian marsupials. They are evolved differently due to different in climatic and geographic regions.

"The wolf (placental mammal) and Tasmanian wolf (marsupial) show convergent evolution, developing similar traits due to similar environments, not shared ancestry."

The centromere of each chromosome divides into two so that there is one centromere for each chromatid in Anaphase II. It is the stage when sister chromatids of every chromosome separate and begin to move towards the opposite ends of the cell. The separation and the movement is due to the shortening of the kinetochore microtubules. Anaphase II precedes telophase II.

The sequence of the 5’ and 3’ primers for the given DNA sequence will be 5’ AATGC 3’, 5’ GACTC 3’. In primer design, the forward primer is derived directly from the start of the DNA sequence, which is 5’ AATGC 3’. The reverse primer is complementary to the end of the sequence, considering the antiparallel nature of DNA, resulting in 5’ GACTC 3’. This follows the standard method for designing primers, as discussed in PCR techniques and DNA replication processes.

Explanation of Assertion: Biofertilizers are living microorganisms that, when applied to seeds, plant surfaces, or soil, colonize the rhizosphere or the interior of the plant and promote growth by increasing the supply or availability of primary nutrients to the host plant. They are a natural and environmentally friendly way to provide essential nutrients to plants, which in turn boosts soil fertility and productivity.

Explanation of Reason: Organic farming is an agricultural system that seeks to provide the consumer with fresh, tasty, and authentic food while respecting natural life-cycle systems. It involves the use of natural fertilizers, such as compost, manure, and green manure, instead of synthetic fertilizers. It also relies on natural pest control methods, such as biological control, crop rotation, and the use of resistant plant varieties, instead of chemical pesticides.

While both the assertion and reason are correct statements, the reason does not directly explain the assertion. The assertion is about the role of biofertilizers in enriching soil nutrients, while the reason is about the principles of organic farming, which is a broader concept that encompasses not only the use of biofertilizers but also other natural and sustainable practices.

Topic in NCERT: Microbes as biofertilisers

Line in NCERT: "biofertilisers are organisms that enrich the nutrient quality of the soil."

Escherichia coli and Agrobacterium tumefaciens are the microbes found to be very useful in genetic engineering. 
E. coli is a motile, gram negative, rod shaped bacterium which is a normal inhabitant of human colon. It is most extensively used in bacterial genetics and molecular biology.
Agrobacterium tumefaciens is a soil bacterium. It has Ti plasmid (Tumour inducing plasmid) and it can be used for the transfer of a desired gene in dicot plants.

Line in NCERT: "Agrobacterium tumifaciens, a pathogen of several dicot plants is able to deliver a piece of DNA known as 'T-DNA' to transform normal plant cells into a tumor and direct these tumor cells to produce the chemicals required by the pathogen. Similarly, retroviruses in animals have the ability to transform normal cells into cancerous cells." "When this DNA is transferred into Escherichia coli, a bacterium closely related to Salmonella, it could replicate using the new host's DNA polymerase enzyme and make multiple copies."

Integral membrane proteins are integrated into the membrane. Some stick only partway into the membrane, while others stretch from one side of the membrane to the other and are exposed on either side. These are transmembrane proteins​.

• Physiological barriers to the entry of microorganisms into the human body are tears in the eyes, saliva in the mouth and HCl in the stomach.
• The enzymes lysozymes are found in tears and saliva and inhibit the synthesis of peptidoglycan present in the cell wall of microorganisms especially eubacteria.
• Disease-causing microorganism enters through different routes into the body. The physiological barrier prevents their entry. Tears act as a physiological barrier for entry of pathogen.

Topic in NCERT: Innate immunity

Line in NCERT: "skin on our body is the main barrier which prevents entry of the micro-organisms."

DPT vaccine is for diphtheria, pertussis and tetanus. The patient suffering from typhoid is required to be treated with antibiotics as it is a bacterial disease.

i: Salmonella typhi infects via contaminated food/water (correct).
ii: Symptoms include high fever, weakness, etc. (correct).
iii: DPT vaccine is for other diseases (incorrect).
iv: Widal test diagnoses typhoid (correct).
v: Antibiotics are required (incorrect).

Answer: Correct (C).

During follicular phase, regeneration of endometrium through proliferation of uterine and fallopian tube cells occurs and as a result, endometrium of uterus becomes considerably thickened in anticipation of pregnancy.

If m is pole strength, then m = M/l
When the wire is bent into a semicircular arc, the separation between the two poles changes from l to 2r, where r is radius of the semicircular arc.
Since l = π r or r = l/π, the new magnetic moment of the steel wire,
M' = m × 2r =

• High input resistance
• Low output resistance
• More efficiency
• More current amplification
• More voltage amplification

Since the charges are placed along the same straight line, the electric field at x = 0 will be directed along the x-axis and its magnitude is given by:



^{which is choice (1).}

Figure shows the field lines (shown as broken curves) of the magnetic field due to the current flowing in the loop. It is clear from the figure that the magnetic flux in the x-y plane will be zero. Hence, the correct choice is (4).

Using conservation of energy
Increase in kinetic energy = Decrease in potential energy K - 0 = q(-dV) ⇒ K = qEy, where q(-dV) is decrease in potential energy. As displacement and electric field are along the same direction, dV = -Ey.

Let the mass of all the liquids be m.
Let s₁, s₂, s₃ be the respective specific heats of the liquids A, B, and C.
The initial temperatures of A, B, and C are 12°C, 19°C, and 28°C, respectively.

Case 1: Mixing A and B
When A and B are mixed, the temperature of the mixture is 16°C.

We know that heat gained by A = heat lost by B,

m s₁ (16 - 12) = ms₂ (19 - 16)

4 s₁ = 3s₂ ...(i)

Case 2: Mixing B and C
When B and C are mixed, the temperature of the mixture is 23°C.

m s₂ (23 - 19) = m s₃ (28 - 23)

4 s₂ = 5 s₃ ...(ii)

From (i) and (ii):
s₁ = (3/4) s₂ = (15/16) s₃

Case 3: Mixing A and C
When A and C are mixed, suppose the temperature of the mixture is t.

m s₁ (t - 12) = m s₃ (28 - t)

Substituting s₁ = (15/16) s₃:

(15/16) s₃ (t - 12) = s₃ (28 - t)

Canceling s₃ from both sides:

(15/16) (t - 12) = (28 - t)

Multiplying by 16:

15t - 180 = 448 - 16t

31t = 628

t = 628 / 31 = 20.3°C

Thus, the final temperature when A and C are mixed is 20.3°C.

Work done by the string in increasing its extension from I1 to I2 is the difference in potential energy stored in it.

We know that the potential energy stored is U = 1/2 Kx².

So, at extension I1,
UI₁ = 1/2 K I₁².

At extension I₂,
UI₂ = 1/2 K I₂².

Work done, W = ΔU = UI₂ - UI₁ = 1/2 K (I₂² - I₁²).

v = 10⁴ m/s
m = 10⁻²⁶ kg
n = 10²² m⁻³
A = 1 m²

Change in momentum, Δp = 2mnv
Δp = 2 × 10²² × 10⁻²⁶ × 10⁴ = 2 N·s

Pressure, P = F/A = 2 N/m² = 2 N m^-2

Sweet makers do not clean the bottom of the cauldron because they need the cauldron to absorb more and more heat and be heated quickly so, they can make sweets in less time.

Black, rough surfaces absorb more heat than bright ones, so sweet makers keep cauldron bottoms unclean to heat quickly.

Correct answer: A

Physical adsorption: Increases with higher pressure, greater surface area, lower temperature (exothermic). Forms multilayers, not unilayer (characteristic of chemisorption).

The second electron affinity refers to the energy change when an electron is added to a negatively charged ion, i.e., when an additional electron is added to an anion.

• Alkali metals (Group 1) typically have a positive electron affinity for the first electron (they form a +1 cation), but the second electron affinity is positive because adding an electron to a negatively charged ion (such as M⁻ → M²⁻) would require energy. Hence, the second electron affinity is not zero for alkali metals.
• Halogens (Group 17) have a negative electron affinity for the first electron (they form a -1 anion), but adding a second electron (to form a -2 anion) would involve repulsion between the two negatively charged ions, making the second electron affinity positive or at least very small, but not zero.
• Noble gases (Group 18) already have a full valence shell, and their second electron affinity is zero because they already have a stable electron configuration, and adding an electron would disrupt this stable configuration. Therefore, the second electron affinity is zero for noble gases.
• Transition metals have varying electron affinities, but generally, their second electron affinity is not zero. These metals do not exhibit a simple trend like noble gases.

Conclusion: The second electron affinity is zero for Noble gases.
Thus, the correct answer is: C: Noble gases.

Let the cost price of an article be ₹x.

Selling price of the article at a gain of 12.5% is:
SP = Cost Price + Gain % of Cost Price
⇒ x + 12.5% of x
⇒ x + (1/8) × x
⇒ ₹(9/8) x

Selling price of the article at a gain of 25% is:
SP = Cost Price + Gain % of Cost Price
⇒ x + 25% of x
⇒ x + (1/4) × x
⇒ ₹(5/4) x

According to the question, the difference between the selling prices at 25% and 12.5% gain is ₹22.50:
(5/4)x - (9/8)x = ₹22.50

Converting to a common denominator:
(10x/8) - (9x/8) = ₹22.50
(x/8) = ₹22.50
x = ₹22.50 × 8
x = ₹180

Thus, the cost price of the article is ₹180.